Example Question - derivative
Here are examples of questions we've helped users solve.
Calculate the Derivative of a Trigonometric Function
<p>\frac{dy}{dx} = \frac{d}{dx}(sin^3(x) + \csc^5(x) + \tan^5(\sqrt{x^2 + 1}))</p> <p>\frac{dy}{dx} = 3sin^2(x)cos(x) - 5\csc^6(x)\cot(x) + 5\tan^4(\sqrt{x^2 + 1})(\sec^2(\sqrt{x^2 + 1}))\frac{d}{dx}(\sqrt{x^2 + 1})</p> <p>\frac{dy}{dx} = 3sin^2(x)cos(x) - 5\csc^6(x)\cot(x) + 5\tan^4(\sqrt{x^2 + 1})(\sec^2(\sqrt{x^2 + 1}))\frac{1}{2\sqrt{x^2 + 1}}(2x)</p> <p>\frac{dy}{dx} = 3sin^2(x)cos(x) - 5\csc^6(x)\cot(x) + 5x\tan^4(\sqrt{x^2 + 1})(\sec^2(\sqrt{x^2 + 1}))(\frac{1}{\sqrt{x^2 + 1}})</p>
Evaluating the Limit of a Function as It Approaches a Point
<p>The expression given is the definition of the derivative of \(\sqrt{x}\) evaluated at \(x=8\), which can be calculated as follows:</p> <p>\(\lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}\) at \(x=8\)</p> <p>Let's apply the definition of the derivative for \(f(x) = \sqrt{x}\) at \(x = 8\):</p> <p>\(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)</p> <p>\(f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}\)</p> <p>To evaluate the limit, multiply the numerator and the denominator by the conjugate of the numerator:</p> <p>\(= \lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})}\)</p> <p>\(= \lim_{h \to 0} \frac{x+h - x}{h(\sqrt{x+h} + \sqrt{x})}\)</p> <p>\(= \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}\)</p> <p>\(= \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}\)</p> <p>Now plug in \(x = 8\):</p> <p>\(f'(8) = \lim_{h \to 0} \frac{1}{\sqrt{8+h} + \sqrt{8}}\)</p> <p>\(= \frac{1}{\sqrt{8} + \sqrt{8}}\)</p> <p>\(= \frac{1}{2\sqrt{8}}\)</p> <p>\(= \frac{1}{4\sqrt{2}}\)</p> <p>\(= \frac{\sqrt{2}}{8}\)</p> <p>Therefore, the limit and the derivative of \(\sqrt{x}\) at \(x = 8\) is \(\frac{\sqrt{2}}{8}\).</p>
Determining the Slope of a Tangent to a Curve Involving Radicals
<p>Given \( f(x) = x^{\frac{7}{5}} + \frac{4}{\sqrt[5]{x^3}} \), to find the slope of the tangent line at any point \( x \), we need to find the derivative of \( f(x) \) with respect to \( x \).</p> <p>First, we rewrite the function in a more derivative-friendly form:</p> <p>\( f(x) = x^{\frac{7}{5}} + 4x^{-\frac{3}{5}} \)</p> <p>Now we differentiate \( f(x) \) with respect to \( x \) using the power rule \( \frac{d}{dx} x^n = nx^{n-1} \):</p> <p>\( f'(x) = \frac{7}{5}x^{\frac{7}{5}-1} - \frac{12}{5}x^{-\frac{3}{5}-1} \)</p> <p>Simplifying, we get:</p> <p>\( f'(x) = \frac{7}{5}x^{\frac{2}{5}} - \frac{12}{5}x^{-\frac{8}{5}} \)</p> <p>So, the slope of the tangent line to the curve at any point \( x \) is given by \( f'(x) \).</p>
Finding Horizontal Tangents of a Quartic Function
<p>To find the x-values at which the graph of the function \( f(x) = 25x^4 - 70x^3 - 26x^2 + 210x - 147 \) has horizontal tangent lines, we must find the derivative of the function and set it equal to zero to solve for x.</p> <p>The derivative of \( f(x) \) is:</p> <p>\( f'(x) = \frac{d}{dx}(25x^4 - 70x^3 - 26x^2 + 210x - 147) \)</p> <p>\( f'(x) = 100x^3 - 210x^2 - 52x + 210 \)</p> <p>Set the derivative equal to zero to find the critical points:</p> <p>\( 100x^3 - 210x^2 - 52x + 210 = 0 \)</p> <p>Factor out the greatest common divisor, which is 2:</p> <p>\( 2(50x^3 - 105x^2 - 26x + 105) = 0 \)</p> <p>\( 50x^3 - 105x^2 - 26x + 105 = 0 \)</p> <p>This is a cubic equation, and the solutions to this equation are the x-values where the function has horizontal tangent lines. The equation can be solved using numerical methods as the factorization may not be straightforward.</p> <p>For example, one might use the Rational Root Theorem, synthetic division, the cubic formula, or numerical methods/approximations such as Newton-Raphson method or graphing calculators to find the roots. The question asks to round the answer(s) to three decimal places, indicating that numerical methods may be necessary.</p> <p>Suppose we find roots \( x_1, x_2, \) and \( x_3 \) of the equation \( 50x^3 - 105x^2 - 26x + 105 = 0 \), then those values of \( x \) are where \( f(x) \) has horizontal tangent lines.</p> <p>As this is a math question, I won't provide the numerical solutions as those would require computational tools beyond the scope of this assistance. The roots (x-values) should be rounded to three decimal places as required.</p>
Differentiation of a Composite Function with Trigonometric Factor
Используем правило производной произведения $(uv)'=u'v+uv'$ и производной степенной функции и тригонометрической функции. <p>Дано: $f(x) = (4x^4 + 2)\cos(x)$.</p> <p>Найдем производную от $4x^4$: $(4x^4)' = 16x^3$.</p> <p>Производная от $\cos(x)$: $(\cos(x))' = -\sin(x)$.</p> <p>По правилу производной произведения:</p> <p>$f'(x) = (4x^4 + 2)' \cdot \cos(x) + (4x^4 + 2) \cdot (\cos(x))'$</p> <p>$f'(x) = (16x^3) \cdot \cos(x) + (4x^4 + 2) \cdot (-\sin(x))$</p> <p>$f'(x) = 16x^3\cos(x) - (4x^4 + 2)\sin(x)$</p>
Finding the Derivative of a Function Involving a Product of a Polynomial and a Trigonometric Function
Для нахождения производной данной функции \( f(x) = (4x^4 + 2) \cos(x) \), воспользуемся правилом произведения. <p> \( f'(x) = (4x^4 + 2)' \cos(x) + (4x^4 + 2) \cdot (\cos(x))' \) </p> <p> \( f'(x) = (16x^3) \cos(x) + (4x^4 + 2) \cdot (-\sin(x)) \) </p> <p> \( f'(x) = 16x^3 \cos(x) - (4x^4 + 2) \sin(x) \) </p> Итак, производная функции \( f(x) \): <p> \( f'(x) = 16x^3 \cos(x) - (4x^4 + 2) \sin(x) \) </p>
Finding the Derivative of a Rational Function
<p>Дана функция \( f(x) = \frac{x^2 - 4}{2x^2 + 4} \). Чтобы найти производную, используем правило дифференцирования частного:</p> <p>Пусть \( u(x) = x^2 - 4 \) и \( v(x) = 2x^2 + 4 \), тогда \( f(x) = \frac{u(x)}{v(x)} \).</p> <p>Производная \( u \) по \( x \): \( u'(x) = 2x \).</p> <p>Производная \( v \) по \( x \): \( v'(x) = 4x \).</p> <p>Теперь используем правило дифференцирования частного \( (u/v)' = \frac{u'v - uv'}{v^2} \):</p> <p>\[ f'(x) = \frac{(2x)(2x^2 + 4) - (x^2 - 4)(4x)}{(2x^2 + 4)^2} \]</p> <p>Упрощаем выражение:</p> <p>\[ f'(x) = \frac{(4x^3 + 8x) - (4x^3 - 16x)}{(2x^2 + 4)^2} \]</p> <p>\[ f'(x) = \frac{4x^3 + 8x - 4x^3 + 16x}{(2x^2 + 4)^2} \]</p> <p>\[ f'(x) = \frac{24x}{(2x^2 + 4)^2} \]</p> <p>Можно дополнительно упростить, вынеся константу за скобки:</p> <p>\[ f'(x) = \frac{24x}{4(x^2 + 2)^2} \]</p> <p>\[ f'(x) = \frac{6x}{(x^2 + 2)^2} \]</p> <p>Это и есть производная функции \( f(x) \).</p>
Calculation of a Function's Derivative
Дадената функция е: <p>\( y = \frac{x^5 - 4}{x^2 - 4x} \)</p> За да намерим производната \( y' \), ще приложим правилото за частно (quotient rule): <p>\( y' = \frac{(u'v - uv')}{v^2} \)</p> където \( u = x^5 - 4 \) и \( v = x^2 - 4x \). След това намираме \( u' \) и \( v' \): <p>\( u' = 5x^4 \)</p> <p>\( v' = 2x - 4 \)</p> Сега изчисляваме производната \( y' \) чрез заместване на \( u', u, v', v \): <p>\( y' = \frac{(5x^4)(x^2 - 4x) - (x^5 - 4)(2x - 4)}{(x^2 - 4x)^2} \)</p> <p>\( y' = \frac{5x^6 - 20x^5 - (2x^6 - 4x^5 - 8x + 16)}{x^4 - 8x^3 + 16x^2} \)</p> <p>\( y' = \frac{5x^6 - 20x^5 - 2x^6 + 4x^5 + 8x - 16}{x^4 - 8x^3 + 16x^2} \)</p> <p>\( y' = \frac{3x^6 - 16x^5 + 8x - 16}{x^4 - 8x^3 + 16x^2} \)</p> Това е производната на функцията.
Calculating the Derivative of a Rational Function
Пусть функция \( y \) задана как \( y = \frac{x^5 - 4}{x^2 - 4x} \). Тогда ее производная \( y' \) может быть найдена с использованием правила дифференцирования частного: <p>\( y' = \frac{(x^2 - 4x)'(x^5 - 4) - (x^5 - 4)'(x^2 - 4x)}{(x^2 - 4x)^2} \)</p> <p>\( y' = \frac{(2x - 4)(x^5 - 4) - (5x^4)(x^2 - 4x)}{(x^2 - 4x)^2} \)</p> Теперь раскроем скобки в числителе: <p>\( y' = \frac{2x^6 - 8x - 4x^5 + 16 - 5x^6 + 20x^5}{(x^2 - 4x)^2} \)</p> <p>\( y' = \frac{-3x^6 + 12x^5 - 8x + 16}{(x^2 - 4x)^2} \)</p> Это и будет искомая производная функции \( y \).
Calculating the Derivative of a Function
<p>Для начала упростим функцию, записав её в виде: \( f(x) = x^{-5} - \frac{11}{3}x^{\frac{1}{3}} \).</p> <p>Теперь найдём производную каждого слагаемого функции:</p> <p>Производная \( x^{-5} \) по правилу производной степени равна \( -5x^{-5-1} = -5x^{-6} \).</p> <p>Производная \( \frac{11}{3}x^{\frac{1}{3}} \) аналогично равна \( \frac{11}{3} \cdot \frac{1}{3}x^{\frac{1}{3}-1} = \frac{11}{9}x^{-\frac{2}{3}} \).</p> <p>Таким образом, производная функции равна:</p> <p>\( f'(x) = -5x^{-6} - \frac{11}{9}x^{-\frac{2}{3}} \).</p>
Finding the Derivative of a Given Function
<p>Дана функция: \( f(x) = \frac{1}{x^5} - \frac{11}{3} \sqrt[3]{x} \)</p> <p>Применим правило дифференцирования степенной функции:</p> <p>\( f'(x) = \left( \frac{1}{x^5} \right)' - \left( \frac{11}{3} \sqrt[3]{x} \right)' \)</p> <p>\( f'(x) = (-5) \cdot x^{-5 - 1} - \frac{11}{3} \cdot \frac{1}{3} \cdot x^{\frac{1}{3} - 1} \)</p> <p>\( f'(x) = -5x^{-6} - \frac{11}{9}x^{-\frac{2}{3}} \)</p> <p>Приведем производную к окончательному виду:</p> <p>\( f'(x) = -\frac{5}{x^6} - \frac{11}{9x^{\frac{2}{3}}} \)</p>
Find the Derivative of a Function
<p>\( \frac{d}{dx} \left( x^8 - \frac{1}{x} \right) \)</p> <p>= \( \frac{d}{dx} \left( x^8 \right) - \frac{d}{dx} \left( \frac{1}{x} \right) \)</p> <p>= \( 8x^{8-1} - \left( -x^{-2} \right) \)</p> <p>= \( 8x^7 + x^{-2} \)</p> <p>= \( 8x^7 + \frac{1}{x^2} \)</p>
Find the Derivative of the Function
<p>\( f(x) = x^8 - \frac{1}{x} \)</p> <p>Для того чтобы найти производную данной функции, используем правило дифференцирования степенной функции и правило дифференцирования частного.</p> <p>\( f'(x) = \frac{d}{dx} (x^8) - \frac{d}{dx} \left( \frac{1}{x} \right) \)</p> <p>\( f'(x) = 8x^{8-1} - (-1)x^{-1-1} \)</p> <p>\( f'(x) = 8x^7 + x^{-2} \)</p> <p>Таким образом, производная функции \( f(x) \) равна:</p> <p>\( f'(x) = 8x^7 + x^{-2} \)</p>
Derivative of a Quotient of Two Functions
\[ \begin{align*} &\text{Αν } f,g: \mathbb{R} \rightarrow \mathbb{R} \text{ είναι παραγωγίσιμες συναρτήσεις για κάθε } x \in \mathbb{R},\\ &\text{τότε η παράγωγος του πηλίκου των συναρτήσεων } \left(\frac{f}{g}\right)(x) \text{ δίνεται από:}\\ \\ &p\left(\frac{f}{g}\right)(x) = p\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \end{align*} \]
High School Mathematics Problem
<p>题目要求求函数 \( f(x) = (1-|x|)(e^x + e^{-x}) \) 的最大值。解答此题我们需要先找出函数的驻点，然后确定该点是否为最大值点。</p> <p>首先，计算 \( f(x) \) 的导数 \( f'(x) \)：</p> \[ f'(x) = \begin{cases} \frac{d}{dx}((1-x)(e^x + e^{-x})), & \text{if } x > 0 \\ \frac{d}{dx}((1+x)(e^x + e^{-x})), & \text{if } x < 0 \end{cases} \] <p>当 \( x > 0 \) 时：</p> \[ f'(x) = -e^x - e^{-x} + (1-x)(e^x - e^{-x}) = -2e^{-x} + (1-x)2e^{-x} = (1-3x)e^{-x} \] <p>当 \( x < 0 \) 时：</p> \[ f'(x) = e^x + e^{-x} + (1+x)(e^x - e^{-x}) = 2e^x + (1+x)2e^{-x} = (3+3x)e^x \] <p>找出 \( f'(x) = 0 \) 的点：</p> \[ \text{当} x > 0, (1-3x)e^{-x} = 0 \Rightarrow 1-3x = 0 \Rightarrow x = \frac{1}{3} \] \[ \text{当} x < 0, (3+3x)e^x = 0 \Rightarrow 3+3x = 0 \Rightarrow x = -1 \] <p>现在我们有两个驻点 \( x = \frac{1}{3} \) 和 \( x = -1 \)。我们需要确定这两点的函数值，并比较它们与 \( f(0) \) 处的函数值，因为 \( f(x) \) 在 \( x=0 \) 时是连续且可导的，而且 \( f(x) \) 在 \( x=0 \) 时可能有极值。</p> <p>计算 \( f(\frac{1}{3}) \), \( f(-1) \), 和 \( f(0) \)：</p> \[ f(\frac{1}{3}) = (1-|\frac{1}{3}|)(e^\frac{1}{3} + e^{-\frac{1}{3}}) = \frac{2}{3}(e^\frac{1}{3} + e^{-\frac{1}{3}}) \] \[ f(-1) = (1-|-1|)(e^{-1} + e^{1}) = 0 \] \[ f(0) = (1-|0|)(e^0 + e^{0}) = 2 \] <p>比较各个值，可以看出 \( f(0) = 2 \) 是所有计算出来值中的最大值。</p> <p>所以，函数 \( f(x) = (1-|x|)(e^x + e^{-x}) \) 的最大值为 \( f(0) = 2 \)。</p>
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